Question: Find the area in the plane contained by the graph of
\[|x + y| + |x - y| \le 4.\]
Solution: First, assume that $x \ge 0$ and $y \ge 0.$  If $y \ge x,$ then
\[|x + y| + |x - y| = x + y + y - x = 2y \le 4,\]so $y \le 2.$  If $y < x,$ then
\[|x + y| + |x - y| = x + y + x - y = 2x \le 4,\]so $x \le 2.$

Thus, the portion of the graph in the first quadrant is as follows:

[asy]
unitsize (1 cm);

fill((0,0)--(2,0)--(2,2)--(0,2)--cycle,gray(0.7));
draw((2,0)--(2,2)--(0,2));
draw((-0.5,0)--(2.5,0));
draw((0,-0.5)--(0,2.5));

dot("$2$", (2,0), S);
dot("$2$", (0,2), W);
[/asy]

Now, suppose $(a,b)$ satisfies $|x + y| + |x - y| \le 4,$ so
\[|a + b| + |a - b| \le 4.\]If we plug in $x = a$ and $y = -b,$ then
\[|x + y| + |x - y| = |a - b| + |a + b| \le 4.\]This means if $(a,b)$ is a point in the region, so is $(a,-b).$  Therefore, the region is symmetric around the $x$-axis.

Similarly, if we plug in $x = -a$ and $y = b,$ then
\[|x + y| + |x - y| = |-a + b| + |-a - b| = |a - b| + |a + b| \le 4.\]This means $(-a,b)$ is also a point in the region.  Therefore, the region is symmetric around the $y$-axis.

We conclude that the whole region is a square with side length 4.

[asy]
unitsize (1 cm);

filldraw((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,gray(0.7));
draw((-2.5,0)--(2.5,0));
draw((0,-2.5)--(0,2.5));

dot("$2$", (2,0), SE);
dot("$2$", (0,2), NW);
dot("$-2$", (-2,0), SW);
dot("$-2$", (0,-2), SW);
[/asy]

Hence, its area is $\boxed{16}.$